- Home
- Standard 11
- Mathematics
Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .
$76$
$68$
$82$
$56$
Solution
Let number be $a _{1}, a _{2}, a _{3}, \ldots \ldots a _{2 n }, b _{1}, b _{2}, b _{3} \ldots b _{ n }$
$\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3 n}-(5)^{2}$
$\Rightarrow \sum a^{2}+\sum b^{2}=87 n$
Now, distribution becomes
$a _{1}+1, a _{2}+1, a _{3}+1, \ldots \ldots a _{2 n }+1, b _{1}-1,b_{2}-1 \ldots \ldots b_{n}-1$
Variance
$=\frac{\sum(a+1)^{2}+\sum(b-1)^{2}}{3 n}-\left(\frac{12 n+2 n+3 n-n}{3 n}\right)^{2}$
$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}$
$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}-\left(\frac{16}{3}\right)^{2}$
$=\frac{87 n+3 n+2(12 n)-2(3 n)}{3 n}-\left(\frac{16}{3}\right)^{2}$
$\Rightarrow k=\frac{108}{3}-\left(\frac{16}{5}\right)^{2}$
$\Rightarrow 9 k=3(108)-(16)^{2}=324-256=68$
